Factoring Radicals - Algebra II

First, factor the coefficient:

It's definitely not obvious that 13 goes into 117 unless you examine the potential answers.

We can see that we have a group of two 2's, and a group of two 3's, so those will be on the outside of the radical. We also have 13 leftover.

Now we can look at the variables. We have which can easily be re-written as . This means that when we take the square root we'll get .

We have which can almost just as easily be written as , which is helpful because 90 is an even number. That can be re-written as , so when we take the square root we will get with one leftover.

Putting this all together gives:

To simplify a square root, factor out perfect squares:

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Find the factors of 128 to simplify the term.

We can rewrite the expression as the square roots of these factors.

Simplify.

Start by finding factors for the radical term.

We can rewrite the radical using these factors.

Simplify the first term.

Use the multiplication property of radicals to split the fourth roots as follows:

Simplify the new roots:

Use the multiplication property of radicals to split the perfect squares as follows:

Simplify roots,

In order to solve this equation, we must see how many perfect cubes we can simplify in each radical.

First, let's simplify the coefficient under the radical. is the perfect cube of . Therefore, we can remove from under the radical, and what we have instead is:

Now, in order to remove variables from underneath the square root symbol, we need to remove the variables by the cube. Since radicals have the property

we can see that

With the expression in this form, it is much easier to see that we can remove one cube from , two cubes from , and two cubes from , and therefore our solution is:

To simplify radicals, we need to factor the expression inside the radical. A radical can only be simplified if one of the factors has a square root that is an integer.

For this problem, we'll first find all of the possible radicals of 12: 1 & 12, 2 & 6, and 3 & 4. Then we look at each factor and determine if any of them has a square root that is an integer. The only one that does is 4, which has a square root of 2. We can rewrite the radical as which can also be written as . Taking the squareroot of 4, we come to the answer: .

The goal of simplifying a radical by factoring is to find a factor of the radicand that has a neat whole number as a square root. If this factor is difficult to determine simply by looking at the radicand, a good way to start is by factoring the radicand until you notice a factor that has a whole number as a square root, and can therefore be taken out of the radical. Assuming we couldn't identify a factor of 150 with a neat square root, we could start the factoring by taking out the smallest factor possible, which for even numbers would be 2:

Neither number has a neat square root, so we'll continue by factoring out the next smallest factor of 3 from 75:

At this point we can see that one of our factors, 25, has a neat square root of 5, which we can take out from under the radical, and now that none of the factors left under the radical can be simplified any further, we simply multiply them back together to give us the most simplified form of our radical:

In order to simplify each radical, we must find the factors of its radicand that have a whole number as a square root, which will allow us to take the square root of that factor out of the radical. We start by factoring each radicand, looking for any factors that have a neat whole number as a square root:

After factoring each radicand, we can see that there is a perfect square in each: 25 in the first, 49 in the second, and 4 in the third. Because these factors are perfect squares, we can easily take their square root out of the radical, which then gets multiplied by the coefficient already in front of the radical:

After simplifying each radical, we're left with the same value of in each term, so we can now add all of our like terms together to completely simplify the expression:

When simplifying radicals, you want to factor the radicand and look for square numbers.

Both the and are not perfect squares, so the answer is just .

When simplifying radicals, you want to factor the radicand and look for square numbers.

is a perfect square so the answer is just .

When simplifying radicals, you want to factor the radicand and look for square numbers.

If you aren't sure whether can be factored, use divisibility rules. Since it's odd and not ending in , lets check if is a possible factor. To know if a number is factorable by , you add the sum of the individual numbers in that number.

Since is divisible by , just divide by and continue doing this. You should do this more times and see that .

Since is a perfect square, or we can simplify the radical as follows.

When simplifying radicals, you want to factor the radicand and look for square numbers.

Since is a perfect square but is not, we can write it like this:

Remember, the other factor that's not a perfect square is left in the radicand and the square root of the perfect square is outside the radical.

When simplifying radicals, you want to factor the radicand and look for square numbers.

Since is a perfect square but is not, we can write it as follows:

Remember, the other factor that's not a perfect square is left in the radicand and the square root of the perfect square is outside the radical.

When simplifying radicals, you want to factor the radicand and look for square numbers.

Since is a perfect square but is not, we can write it like this:

Remember, the other factor that's not a perfect square is left in the radicand and the square root of the perfect square is outside the radical.

If you didn't know was a perfect square, there's a divisibility rule for . If the ones digit and hundreds digit adds up to the tens value, then it's divisibile by . so is divisible by and its other factor is also . It's best to memorize perfect squares up to

When simplifying radicals, you want to factor the radicand and look for square numbers.

It's essentially the same factor so the answer is .

Let's see if this quadratic can be broken down. Remember, we need to find two terms that are factors of the c term that add up to the b term.

It turns out that .

The second power and square root cancel out leaving you with just .

Let's factor a out first since they divide evenly with all of the terms.

We get

Let's see if this quadratic can be broken down. Remember, we need to find two terms that are factors of the c term that add up to the b term.

It turns out that .

We have perfect squares so the final answer is .